Answer to Question #293960 in Discrete Mathematics for Tege

Question

Determine the cofficent of X⁵Y¹⁰Z⁵W⁵ in (x-7y+3z-25)²⁵

Answer 1

Since the term $(x - 7y +3z - 25)^{25}$ do not contain $w$, the coefficient of $x^{5}y^{10}z^{5}w^{5}$ must be zero.

So we consider $(x - 7y +3z - 2w)^{25}$ instead of the original problem $(x - 7y +3z - 25)^{25}$ (assuming it a typo error).

We know that by the Multinomial theorem,

$(x - 7y +3z - 2w)^{25} = \displaystyle\sum_{r_1 + r_2 + r_3 + r_4=25}\dfrac{25!}{r_1!r_2!r_3!r_4!}(x)^{r_1}(-7y)^{r_{2}}(3z)^{r_{3}}(-2w)^{r_{4}}$

The coefficient of $x^{5} y^{10}z^{5}w^{5}$ in $(x - 7y +3z - 2w)^{25}$ is the term obtained by taking $r_1 = 5, r_2 = 10, r_3 = 5, r_4 = 5$ in the above summation.

Hence, the coefficient of $x^{5}y^{10}z^{5}w^{5}$

$\begin{aligned} &= \dfrac{25!}{5!10!5!5!}(1)^{5}(-7)^{10}(3)^{5}(-2)^{5}\\ &=-\dfrac{25!}{5!10!5!5!}(7)^{10}(3)^{5}(2)^{5}\\ \end{aligned}$